 | Isaac Todhunter - Geometry - 1855 - 332 pages
...lines drawn from the angles of a triangle to the middle points of the opposite sides meet in a point. Let ABC be a triangle, D, E, F, the middle points of the sides ; take A for the origin, AB for the direction of the axis of x, and a line through A perpendicular... | |
 | Isaac Todhunter - Conic sections - 1862 - 342 pages
...lines drawn from the angles of a triangle to the middle points of the opposite sides meet in a point. Let ABC be a triangle, D, E, F the middle points of the sides ; take A for the origin, AB for the direction of the axis of x, and a line through A perpendicular... | |
 | Robert Hudson Graham - Graphic statics - 1883 - 478 pages
...line, joining these two points, must be parallel to the line of action of the resultant force. 13. Let, ABC, be a triangle, D, E, F, the middle points of the sides opposite to A, B, and C, respectively; shew, by aid of the polar and force polygons, that... | |
 | Robert Hudson Graham - Graphic statics - 1883 - 438 pages
...line, joining these two points, must be parallel to the line of action of the resultant force. 13. Let, ABC, be a triangle, D, E, F, the middle points of the sides opposite to A, B, and C, respectively ; shew, by aid of the polar and force polygons, that... | |
 | Henry Martyn Taylor - 1893 - 486 pages
...is double of thc straight line joining the mtddle points of the other sides and is parallel to it. Let ABC be a triangle, D, E, F the middle points of the sides BC, CA, AB. Draw BE, CF, EF, FD, DE. BDG Because the two triangles BFC, AFC have their sides... | |
 | Thomas Grenfell Vyvyan - Geometry - 1894 - 178 pages
...the opposite sides, (ii) perpendicular to the sides, (iii) bisecting the angles, are concurrent. i. Let ABC be a triangle, D, E, F the middle points of the sides, and let the coordinates of А, В, С be (haks) respectively, any axes being taken, then... | |
 | Henry Martyn Taylor - Euclid's Elements - 1895 - 708 pages
...is double of the straight line joining the middle points of the other sides and is parallel to it. Let ABC be a triangle, D, E, F the middle points of the siilos BC, CA, AB. Draw BE, CF, EF, FD, DE. E B vc Because the two triangles BFC, AFC have their... | |
 | h. m. taylor - 698 pages
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Because the two triangles BFC, AFC have their sides BF, AF equal and in a straight line, and the point C common, the triangles are equal; (Prop. 38, Coroll.) therefore the triangle BFC is half of the triangle ABC. Similarly it can be proved that the... 